# Thread: I got some crazy simultaneous equation that no one can solve

1. ## I got some crazy simultaneous equation that no one can solve

Here it is:

$(a + b)x + cy = bc$

$(b + c)y + ax = -ab$

Find x and y.

I've tried this for ages but the answers are $x= c$ and $y=-a$

2. In matrix form this would be

$\displaystyle \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}bc\\8722ab\end{matrix}\right]$

$\displaystyle \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]^{-1}\left[\begin{matrix}bc\\8722ab\end{matrix}\right]$.

Can you go from here?

3. By the substitution method:

$[1] \,\,\, (a + b)x + cy = bc$

$[2] \,\,\, (b + c)y + ax = -ab$

$[2] \,\,\, y = \dfrac{-ab - ax}{b + c}$

Substitute into [1]:

$[1] \,\,\, (a + b)x + \dfrac{c(-ab - ax)}{b + c} = bc$

And the simplify to solve for x. Place the value of x back into one of the equations to solve for y.

4. Hello, Prove it!

Be careful . . . The original equations were "scrambled" . . .

$\begin{array}{ccc}(a + b)x + cy &=& bc \\
ax + (b + c)y &=& \text{-}ab\end{array}\qquad \text{(unscrambled form)}$

We want: . $\begin{bmatrix}a+b & c \\ a & b+c\end{bmatrix}\,\begin{bmatrix} x \\ y\end{bmatrix} \;=\;\begin{bmatrix}bc \\ \text{-}ab\end{bmatrix}$

5. WOW!!! I GOT IT USING MATRICES!!!!!!!

WOOOOOOOOOOOOOOOOOOOOOOOO

Thanks Prove It and Soroban

I have too much working to type it all but the determinant is $(a+b)(b+c)-ac$ Then just simplify away.

thanks!!!!

6. Originally Posted by jgv115
$(a + b)x + cy = bc$
$(b + c)y + ax = -ab$
I've tried this for ages but the answers are $x= c$ and $y=-a$
Huh? Try a = 2, b = 4, c = -6, x = 1, y = 5

That's a solution for both equations.....but x<>c and y<>-a
There are many other similar cases.

Did I miss something?

Also, equations can be re-arranged this way:
ax + bx + cy = bc [1]
-ax - by - cy = ab [2]
===============
bx - by = ab + bc
x - y = a + c